Conjecture: Let $A(m,n)$ denote the number of regions in $\mathbb{R}^m$ formed by $n$ hyperplanes and $T_{m,n}$ denote the $n$th term of a sequence whose $(m-1)$th differences are $1,2,3,4,5,…$. It then holds that $T_{m,n} = A(m, n-1)$.
My question is motivated by my answer to this question where I show that if the above conjecture is true, then $T_{m,k} = \sum_{k=0}^{m} {n-1 \choose k} \tag{1}$
It follows that the converse is true i.e. $(1) \implies$ the above conjecture is true. The user QC_QAOA has now proven $(1)$ which proves the above conjecture. However, I would like to know a way to prove directly that $T_{m,n} = A(m, n-1)$ without having to prove $(1)$.
My attempt was as follows (quoted from my original answer):
Note that $$A(m, n) = A(m, n-1) + A(m-1, n-1) \iff A(m,n) – A(m, n-1) = A(m-1, n-1)$$
(Proof here)
If Claim $1$ is true, it follows that $$T_{m, n} = A(m, n-1) \implies T_{m, n+1} = A(m, n)$$ $$T_{m, n+1} – T_{m, n} = A(m, n) – A(m, n-1) = A(m-1, n-1) \tag{1}$$
I have been unable to prove $(1)$, i.e. the difference between the
$(n+1)$th and $n$th terms of a sequence whose $(m-1)$th differences are $1,2,3,4,5…$
is the same as the number of regions formed by $n-1$ hyperplanes in
$\mathbb{R}^{m-1}$. If someone is able to demonstrate this, however,
the proof is complete.
Hence, I was able to transform the problem into proving $T_{m, n+1} – T_{m, n} = A(m-1, n-1)$. However, I am not sure how to take this forward.
Best Answer
The answer here has exactly what you need. They show the formula $$A(m,n) = A(m, n-1) + A(m-1, n-1)$$ By considering removing and adding a hyperplane, in particular noting that the number of regions added will be the number of regions the other hyperplanes form on the new hyperplane.
Thus, writing $A(m,n) = A(m)(n)$we get that $$\Delta [A(m)](n) =A(m)(n+1)-A(m)(n) = A(m-1)(n)$$ And so by induction $$\Delta^{m-1}[A(m)](n) = A(1)(n) = n+1$$ which is how you've defined $T_{m,n+1}$, so we're done.